I'm trying to understand something theoretical about this question...
There's this ball which falls from height $h$ into a bowl, there are $3$ named points on the bowl, $A$ the first upper point the ball hits, $C$ same as $A$ but on the opposite direction, and $B$ which is at the bottom of the bowl.
There's no friction.
In the horizontal axis, there's momentum conservation because no external forces (also there's energy conservation, all forces are conservative, no friction...).
I understand that at points $A$ and $C$ the velocities are equal since both objects are moving together (horizontally only), and at point $B$ they're not.
But how come the answer when finding the velocities at point C is as follows:
(also, initial horizontal momentum is zero, since both ball and bowl are not moving horizontally)
(in horizontal axis) $P_i = P_f$ ($i= initial$, $f= final$....this is because momentum conservation)
$(m+M)u = 0$ (both velocities are equal horizontally)
then the conclusion is that the bowl's and the ball's velocity is $0$
Afterwards they used energy conservation doing $(E_i = E_f)$:
$mg(h+R) = mgR + \frac{1}{2}mu^2$ (but this u is for the ball's velocity, I assume vertical velocity, since the horizontal is $0$ as mentioned above)
if both velocities are equal at point $C$ then what's the assumption the bowl's velocity vertically is $0$ and the ball's is not (as used in the energy part. Obviously the ball is not since it's moving on the bowl)? could someone please help me understand what I'm missing \ misunderstanding? I can't figure out what I'm missing in the answer for which they only used the velocity of the ball in the energy conservation equation
(*still didn't learn center of mass in here, so can't depend on it)
a drawing I made to help understand where each point is...
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$\begingroup$ Are you asking why you can assume the vertical velocity of the bowl at point C is zero? The fact you know all the momentum is vertical and the fact that the movement of the bowl is constrained in the vertical direction by the table seems enough to me. (Also, we are we assuming this bowl us curved on the outside as drawn? Or is that curve just the inside and the outside is actually flat bottom so can't rock?) $\endgroup$– DKNguyenMay 27, 2022 at 18:45
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1$\begingroup$ I mean, isn't it enough just to know that $v_y = 0$? It's like problems where there are rollers on horizontal or vertical guide rails. You know certain values in certain directions are not allowed so you fix those at zero. It's not unlike how you might use a cosine or sine to relate one the $v_x$ to $v_y$ on a guide rail that is being held at an angle. I would be a lot more confused about how you know the entry of the ball doesn't cause a bounce. $\endgroup$– DKNguyenMay 27, 2022 at 19:06
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2$\begingroup$ Well, you would know there is table there (I'm assuming that's specified somewhere in the problem) and you would probably assume that the ball doesn't smash through the bowl or the table. $\endgroup$– DKNguyenMay 27, 2022 at 19:07
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1$\begingroup$ I don't know how to tell if it bounces or not though. I assume anything other than a perfect half circle where the ball doesn't follow the tangent upon entry causes a bounce but I don't know how you differentiate that from the normal force pointing away from the surface while it is in the arc vs just entering the arc. $\endgroup$– DKNguyenMay 27, 2022 at 19:13
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1$\begingroup$ I made a question about bounce here: physics.stackexchange.com/questions/710898/… $\endgroup$– DKNguyenMay 27, 2022 at 19:29
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1 Answer
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I am not sure what your concern is, since your question is ambiguously worded. However, what happens is that the falling ball makes contact with the bowl, and as it moves down the surface of the bowl from A to B it imparts a recoil to the bowl towards the left. Since the bowl is symmetrical, as the ball then rises up the surface from B to C it now imparts an equal recoil to the bowl to the right. Overall the effect on the bowl is zero, so the bowl remains stationary at the end, and all the KE is concentrated in the movement of the ball at C.
answered May 27, 2022 at 19:27
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$\begingroup$ I assumed the recoil happened on the horizontal axis only, but from your answer I see it happens in the whole system of this question (which now sounds more reasonable as well). Just to make sure, did I understand correctly (the recoil happens vertically as well)? $\endgroup$– SA.93May 27, 2022 at 19:35
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1$\begingroup$ Strictly speaking there is a recoil vertically, but the recoil in that direction acts on the bowl, the table and the Earth all together, and since the mass of the Earth is enormously greater than that of the ball, the recoil speed of the Earth is effectively zero. In reality, what happens is not that the Earth as a who recoils, but that the bowl and the table compress slightly, like a spring, then uncompress. $\endgroup$ May 27, 2022 at 19:56